COVENANT UNIVERSITY
NIGERIA
TUTORIAL KIT
OMEGA SEMESTER
PROGRAMME: BUILDING
TECHNOLOGY
COURSE: BLD 224
DISCLAIMER
The contents of this document are intended for practice and leaning purposes at the
undergraduate level. The materials are from different sources including the internet and
the contributors do not in any way claim authorship or ownership of them. The materials
are also not to be used for any commercial purpose.
BLD 224: STRUCTURAL THEORY AND STRENGTH OF MATERIAL II
CONTRIBUTOR: MR OPEYEMI JOSHUA, MR R.A OJELABI
1. A steel bar of rectangular cross-section, 3 cm by 2 cm, carries an axial load of
30 kN. Estimate the average tensile stress over a normal cross-section of the
bar.
2. A cylindrical block is 30 cm long and has a circular cross-section 10 cm in
Diameter. It carries a total compressive load of 70 KN, and under this load it
Contracts by 0.02 cm. Estimate the average compressive stress over a normal
Cross-section and the compressive strain.
3. Explain the following terms: Linear of proportionality, Elastic limit and Breaking point.
4. A steel bolt, 2.50 cm in diameter, carries a tensile load of 40 kN. Estimate the
average tensile stress at the section a and at the screwed section b, where the
diameter at the root of the thread is 2.10 cm.
5.
A tensile test is carried out on a bar of mild steel of diameter 2 cm. The bar
Yields under a load of 80 kN. It reaches a maximum load of 150 KN, and
breaks finally at a load of 70 kN.
Estimate:
(1) The ultimate tensile stress;
(ii) The tensile stress at the yield point;
(iii) The average stress at the breaking point, if the diameter of the
fractured neck is 1 cm.
6
Differentiate between brittle and ductile material and give examples
7
The piston of a hydraulic ram is 40 cm diameter, and the piston rod 6 cm
diameter. The water pressure is 1 MN/m2. Estimate the stress in the piston
rod and the elongation of a length of 1 m of the rod when the piston is under
Pressure from the piston-rod side. Take Young's modulus as E = 200 GN/m2.
8
A bar of cross-section 2.25 cm by 2.25 cm is subjected to an axial pull of
20 kN. Calculate the normal stress and shearing stress on a plane the normal
to which makes an angle of 60o with the axis of the bar, the plane being
Perpendicular to one face of the bar.
9
At a point of a material the two-dimensional stress system is defined by
бx = 60.0 MN/m2, tensile
бy = 45.0 MN/m2, compressive
Ʈxy, = 37.5 MN/m2, shearing.
10. The state of stress at a point in a loaded structure is given by the strength tensile.
140
Tб = 60
0
60
−90
0
0
0
0
Determine
a. The maximum principal stress
b. The minimum principal stress
c. The maximum and minimum shear strain
d. The maximum and minimum shear stress
Taking E = 2.1 x 105N/mm2
J = 0.32
SOLUTION
1.
The area of a normal cross-section of the bar is
Tension and compression: direct stresses
A = 0.03 x 0.02 = 0.6 x l0-3 m2
The average tensile stress over this cross-section is then
Б= P/A
Б= 30 x 103/ 0.6 x 10-3 = 50MN/m2.
2. The area of a normal cross-section is
A = ∏/4 (0.10)2 = 7.85 X 10-3 m2
The average compressive stress over this cross-section is then
Б = P/A = 70 X 103/7.85 X 10-3 = 8.92MN/m2
The average compressive strain over the length of the cylinder is
Ε = 0.02 X 10-2/ 30 X 10-2 = 0.67 X 10-3
5.
The original cross – section of the bar is
A0 = ∏/4 (0.020)2 = 0.314 X 10-3m2
(i)
The average tensile stress at yielding is then
Бy = Py/Ao = 80 X 103/ 0.314 X 10-3 = 254MN/m2
Where Py = load at the yield point.
(ii)
The ultimate stress is the normal stress at the maximum load
Бult = Pmax/ Ao = 150 x 103 / 0.314 x 10-3 = 477MN/m2
Where Pmax is maximum load
(iii)
The cross-sectional area in the fracture neck is
AF = ∏/4 (0.010)2 = 0.0785 X 10-3m2
The average stress at the breaking point is then
Бf = pf/ Af = 70 X 103/ 0.0785 x 10-3 = 892MN/m2
Where Pf = final breaking load.
8. We have θ = 60o, P = 20KN and A = 0.507 x 10 -3 m2. Then
Бx = 20 x 103/ 0.507 x 10-3 = 39.4 MN/m2
The normal stress on the oblique plane is
Б = бx cos260o = (39.4 x 106)1/4 = 9.85MN/m2
The shearing stress on the oblique plan is
½ бx sin 120o = ½ (39.4 x 106) √3/2 = 17.05MN/m2
10.
140
Tб = 60
0
60
−90
0
б𝑥
0
0 = Ʈ𝑦𝑥
Ʈ𝑧𝑥
0
Ʈ𝑥𝑦
б𝑦
Ʈ𝑧𝑦
Ʈ𝑥𝑧
Ʈ𝑦𝑧
б𝑧
Base on the similarity of matrix
бx = 140 N/mm2
Ʈxy = Ʈyx =60 N/mm2
бy = -90 N/mm2
Maximum Principal Stress (б1) = (бx + бy)/2+ 1/2√ (бxx – бyy) 2 + 4Ʈ2xy
Substituting into the equation
Б1 = (140 – 90)/2 + 1/2√ (140 +90)2 + 4 (60)2
25 + ½ √52,900 + 14,400
25+ ½ √67300
25 + ½ X 259.42
25 + 129.71
Б1 = 154.71N/mm2
Minimum Principal Stress (б1) = (бx + бy)/2- 1/2√ (бxx – бyy) 2 + 4Ʈ2xy
Substituting into the equation
Б2 = (140 – 90)/2 + 1/2√ (140 +90)2 + 4 (60)2
25 - ½ √52,900 + 14,400
25- ½ √67300
25 - ½ X 259.42
25 - 129.71
Б2 = -104.71 N/mm2
SHEAR STRESS
Ʈmax = ½ (б1 – б2)
Substituting in the formula
Ʈmax = ½ [(155 – (-105)] =130 N/mm2
Ʈmin = -1/2 [(155 + 105) = -130 N/mm2